🪡 Buffon's needle

What you’re seeing

The floor has lines drawn on it, all the same distance apart. We drop needles at random: a random spot, and a random angle. Some needles cross a line (red), some don’t. Count them, put the counts into a short formula, and you get an estimate of $\pi$. The strip at the bottom shows how the estimate wanders, then closes in on $\pi$ as the needles pile up.

Georges-Louis Leclerc, Comte de Buffon, asked this question in 1733 and published the answer in 1777. It’s one of the first problems in geometric probability.

Why it works

Let the lines be $d$ apart and the needle $\ell$ long, with $\ell \le d$ (so a needle can cross at most one line).

Two numbers describe a drop. Let $y$ be the distance from the needle’s middle to the nearest line, from $0$ to $\tfrac d2$, and let $\theta$ be the angle between the needle and the lines, from $0$ to $\pi$. A fair random drop makes both of them uniform and independent.

When does it cross? The needle reaches $\tfrac\ell2 \sin\theta$ above and below its middle. So it crosses a line exactly when

$$y \le \frac{\ell}{2}\sin\theta.$$

Measure the chance as an area. The pairs $(\theta, y)$ fill a rectangle of area $\pi \cdot \tfrac d2$. The crossing pairs lie under the curve $y = \tfrac\ell2\sin\theta$, which has area

$$\int_0^\pi \frac{\ell}{2}\sin\theta\, d\theta = \frac{\ell}{2}\Big[-\cos\theta\Big]_0^\pi = \ell.$$

So the chance of a crossing is

$$P = \frac{\ell}{\pi d/2} = \frac{2\ell}{\pi d}.$$

Turn it around. If $h$ of $n$ needles cross, then $h/n \approx P$. Solve for $\pi$:

$$\pi \approx \frac{2\ell\, n}{d\, h}.$$

It’s slow. By the law of large numbers, $h/n$ settles down to $P$, but the typical error shrinks only like $1/\sqrt n$. Each extra correct digit needs about 100 times as many needles. Ten thousand drops usually get you two decimal places, sometimes three. That’s why no one computes $\pi$ this way, but it’s a wonderful check that chance and geometry fit together.

A story with a twist. In 1901 Mario Lazzarini reported dropping 3,408 needles and getting $\tfrac{355}{113} = 3.1415929\ldots$, correct to six decimal places. Mathematicians have pointed out that his numbers fit that fraction too neatly: with his needle and line sizes, stopping at just the right moment makes $\tfrac{355}{113}$ come out. A fair run of that size would be very unlikely to do so well.

Try this

• Drop 100 needles a few times (Clear in between). How much do the estimates vary?
• Drop 10,000 several times. How many digits of $\pi = 3.14159\ldots$ do you usually get?
• Make the needles short (0.2 of the gap). Fewer cross, so is the estimate better or worse after 10,000?
• With needle length equal to the gap, the chance of a crossing is $2/\pi \approx 0.637$. Check it against the counts.